Asymptotic Shapes of Large Cells in Random Tessellations

D. Hug and R. Schneider

Abstract. We establish a close relationship between isoperimetric inequalities for con-vex bodies and asymptotic shapes of large random polytopes, which arise as cells incertain random mosaics in d-dimensional Euclidean space. These mosaics are generatedby Poisson hyperplane processes satisfying a few natural assumptions (not necessarily sta-tionarity or isotropy). The size of large cells is measured by a class of general functionals.The main result implies that the asymptotic shapes of large cells are completely deter-mined by the extremal bodies of an inequality of isoperimetric type, which connects thesize functional and the expected number of hyperplanes of the generating process hittinga given convex body. We obtain exponential estimates for the conditional probability oflarge deviations of zero cells from asymptotic or limit shapes, under the condition thatthe cells have large size.

1 Introduction

This paper studies a class of random polytopes and investigates their asymptotic shapes underthe condition that the size of the polytopes becomes large. The polytopes are generated byPoisson processes in the space of hyperplanes of Euclidean space Rd, by taking the cell ofthe induced tessellation of Rd that contains the origin. Our approach includes zero cellsof stationary Poisson hyperplane tessellations as well as typical cells of Poisson–Voronoıtessellations, but goes beyond these special cases. The size of the random polytopes ismeasured by a size functional which is introduced axiomatically. All the common geometricmeasurements of size, like volume, diameter, inradius, and many others, are included. Ashape of a convex body is defined as the orbit of the convex body under a subgroup of thegroup of similarities. If the conditional law for the shape of a random polytope, given the sizeof the random polytope, converges weakly, as the size tends to infinity, to the degenerate lawconcentrated at a fixed shape, then the latter is, by definition, the limit shape of the randompolytope with respect to the chosen size functional. If limit shapes in this sense do not exist,it may still be possible to identify a class of asymptotic shapes, in a weaker sense.

Our main result reveals that the asymptotic shapes of the considered large random poly-topes are completely determined by the extremal bodies of an inequality of isoperimetric typefor convex bodies. It involves two functions on the space of convex bodies: the size func-tional and the expected number of hyperplanes of the defining process hitting a given convexbody. If the extremal bodies of the isoperimetric inequality are unique up to transformationsfrom some group of similarities, then large cells have a corresponding limit shape. Stability

This work was supported in part by the European Network PHD, FP6 Marie Curie Actions, RTN, ContractMCRN-511953.

1

improvements of the isoperimetric inequality yield bounds for the probabilities of large devia-tions from the limit shape. These estimates also provide information on asymptotic shapes incases where limit shapes do not exist. On the other hand, in a special case where the crucialisoperimetric inequality degenerates to a trivial equality, we show that size has no influenceon shape and, hence, limit shapes do not exist.

The topic of shapes of large cells in random tessellations originated from what has becomeknown as D.G. Kendall’s conjecture (see [14], foreword to the first edition). For this, considera stationary and isotropic Poisson line process in the plane and let Z0 be the cell of the inducedrandom tessellation that contains the origin. Kendall’s conjecture stated that the conditionallaw for the shape of Z0, given the area A(Z0) of Z0, converges weakly, as A(Z0) →∞, to thedegenerate law concentrated at the circular shape. A proof was given by Kovalenko [6], [8],who also obtained in [7] an analogous result for the typical cell of a stationary Poisson–Voronoımosaic in the plane. Already before that, Miles [10] had suggested, though not completelyproved, similar results with the area replaced by other functionals, like perimeter, inradius, orwidth in a given direction. Higher-dimensional versions and analogues of Kendall’s conjecturewere investigated in [9], [4], [5]. Of these, [9] treats only a very special case, namely hyperplanemosaics where all cells are parallelepipeds. In [4], Kendall’s problem was extended andsolved for general stationary, not necessarily isotropic Poisson hyperplane processes, withsize measured by the volume. Typical cells of higher-dimensional stationary Poisson–Voronoımosaics were the topic of [5]; here the size was measured by an intrinsic volume.

As a byproduct, we obtain a result on the asymptotic distribution of the size functional,thus extending a result of Goldman [3] from the plane to higher dimensions and to generalsize functionals.

In the next section, we will describe the main results in detail.

2 Assumptions and Main Results

We work in d-dimensional real Euclidean vector space Rd, with scalar product 〈·, ·〉 and norm‖ · ‖. The unit ball x ∈ Rd : ‖x‖ ≤ 1 is denoted by Bd; its boundary is the unit sphereSd−1. By Hd we denote the space of hyperplanes in Rd not containing the origin o, with itsusual topology and Borel structure. For u ∈ Sd−1 and t ∈ R, we write

H(u, t) := x ∈ Rd : 〈x,u〉 = t, H−(u, t) := x ∈ Rd : 〈x,u〉 ≤ t.

Every hyperplane H ∈ Hd has a unique representation H = H(u, t) with u ∈ Sd−1 and t > 0;thus o ∈ H−(u, t). We call this the standard representation. For H ∈ Hd, we denote by H−

the closed halfspace bounded by H that contains o. For a set A ⊂ Hd, we define

P (A) :=⋂

H∈AH−.

Let X be a Poisson hyperplane process in Rd, that is, a Poisson point process in the spaceHd (see [13], for example). We often identify a simple counting measure with its support, sothat both notations, X(A) and card(X ∩A), denote the number of elements of X in A. Weassume that the intensity measure Θ = EX(·) (where E denotes mathematical expectation)is of the form Θ = λµ, with λ > 0 and a measure µ on Hd given by

µ =∫

Sd−1

∫ ∞

01H(u, t) ∈ ·tr−1 dt ϕ(du). (1)

2

Here r ≥ 1, and ϕ is a Borel probability measure on Sd−1 with the property that its supportis not contained in some closed hemisphere. We call λ the intensity and ϕ the directionaldistribution of the hyperplane process X, and to the number r we refer as the distanceexponent.

The random polytopeZ0 := P (X) =

⋂H∈X

H−

is the zero cell, or Crofton cell, of the tessellation induced by X. The case of the zero cell ofa stationary Poisson hyperplane process (of appropriate intensity) is included here, for r = 1(observe that the hyperplanes of a stationary Poisson hyperplane process almost surely donot contain o). Also included is the case of the typical cell of a stationary Poisson–Voronoımosaic (of appropriate intensity), for r = d.

By Kd we denote the space of convex bodies (nonempty, compact, convex sets) in Rd,equipped with the Hausdorff metric δ, and by Kd

o the subspace of all convex bodies containingthe origin. Our investigation of asymptotic shapes of large zero cells will be governed by threecontinuous homogeneous functionals on the space Kd

o : the parameter, size, and deviationfunctionals. We introduce them now.

For K ∈ Kdo , we define

HK := H ∈ Hd : H ∩K 6= ∅.

We haveEX(HK) = λΦ(K) (2)

withΦ(K) := µ(HK) =

1r

∫Sd−1

h(K, u)rϕ(du), (3)

as follows from (1). Here, h(K, u) = max〈x,u〉 : x ∈ K is the value of the supportfunction of K at u. The function Φ is continuous on Kd

o and homogeneous of degree r, thatis, it satisfies Φ(αK) = αrΦ(K) for α ≥ 0. We call Φ the parameter functional of the processX, since multiplied by the intensity λ, it gives the parameter of the Poisson distribution ofthe random variable X(HK), for K ∈ Kd

o :

P(X(HK) = n) =[Φ(K)λ]n

n!exp−Φ(K)λ

for n ∈ N0; here P denotes the underlying probability.

The size of the zero cell can be measured by any real function Σ on Kdo satisfying the

following natural axioms:

(a) Σ is continuous,(b) not identically zero,(c) homogeneous of some degree k > 0,(d) increasing under set inclusion (K ⊂ M ⇒ Σ(K) ≤ Σ(M)).

Let a function Σ with these properties be given. We call it the size functional. Typicalexamples are volume, surface area, mean width, diameter, thickness, inradius, circumradius,width in a given direction.

It is easy to see (see Section 3) that Φ and Σ satisfy a sharp inequality of isoperimetrictype,

Φ(K) ≥ τ Σ(K)r/k for K ∈ Kdo , (4)

3

with a positive constant τ . That the inequality is sharp means that there exist convex bodiesK ∈ Kd

o with more than one point for which equality holds; every such body is called anextremal body (for given Φ and Σ).

We remark that the extremal bodies have the following probabilistic characterization.Among all convex bodies K ∈ Kd

o of size Σ(K) = 1, precisely the extremal bodies maximizethe probability P(K ⊂ Z0). In fact, suppose that K ∈ Kd

o and Σ(K) = 1. Since X(HK) =0 ⇒ K ⊂ Z0 ⇒ X(HαK) = 0 for all α < 1 and P(X(HK) = 0) = exp−Φ(K)λ, we get

P(K ⊂ Z0) = exp−Φ(K)λ ≤ exp−τΣ(K)r/kλ = e−τλ,

with equality if and only if equality holds in (4).

Our third functional measures the deviation of a convex body from the class of extremalbodies. Again, we introduce it axiomatically. We assume that Φ and Σ are given. A realfunction ϑ on K ∈ Kd

o : Σ(K) > 0 is called a deviation functional if

(a) ϑ is continuous,(b) nonnegative,(c) homogeneous of degree zero,(d) ϑ(K) = 0 for some K ∈ Kd

o with Σ(K) > 0 holds if and only if K is an extremal body.

Such deviation functionals always exist. For example, one could take the canonical deviationfunctional

ϑ(K) :=Φ(K)

τΣ(K)r/k− 1. (5)

However, in concrete examples, the deviation functional should be chosen in such a way thatthe deviation has a simple intuitive geometric meaning, and an inequality ϑ(K) < ε shouldallow an explicit estimate of some geometric distance of K from an extremal body. Onepossibility is to use the Hausdorff metric δ and the diameter D and to define

η(K) := minδ(K, M) : M extremal body/D(K) (6)

for K ∈ Kdo with D(K) > 0. The minimum is attained (see Section 3). Clearly η is a deviation

functional; the homogeneity follows from the fact that together with M also αM for α > 0is an extremal body.

It follows from the properties of the involved functionals that the inequality (4) can bestrengthened to a stability estimate: there exists a continuous function f : R+ → R+ withf(ε) > 0 for ε > 0 and f(0) = 0 such that

Φ(K) ≥ (1 + f(ε))τΣ(K)r/k if ϑ(K) ≥ ε, (7)

for K ∈ Kdo (see Section 3). Any such function f will be called a stability function for Φ,Σ, ϑ.

In concrete cases, explicit stability functions are of interest.

In the following, it will be convenient to assume that every stability function f satisfiesf < 1. This can always be achieved, since a given stability function f may be replaced byminf, 1/2.

After these preparations, we can formulate our main result.

Theorem 1. Suppose that a Poisson hyperplane process X with intensity λ, directionaldistribution ϕ and distance exponent r (which determine the parameter functional Φ), a sizefunctional Σ, a deviation functional ϑ, and a stability function f for Φ,Σ, ϑ as explained are

4

given. With a suitable constant c0 > 0 (depending only on τ), the following holds. If ε > 0and 0 < a < b ≤ ∞, then

P(ϑ(Z0) ≥ ε | Σ(Z0) ∈ [a, b)) ≤ c exp−c0f(ε)ar/kλ

, (8)

where c is a constant depending only on ϕ, r,Σ, f, ε.

The implications of this theorem on the existence and nature of limit shapes and asymptoticshapes in general and in various concrete examples will be discussed in Sections 4 and 5. Theproof of Theorem 1 follows in Section 6.

Note that the intensity λ, not appearing explicitly on the left-hand side of (8), togetherwith ϕ and r determines the distribution of the zero cell Z0. We could restrict ourselves toprocesses with intensity one, but would then loose the information about the form of theessential parameter ar/kλ.

In [5, Theorem 2], the special case of Theorem 1 was treated where the Poisson hyperplaneprocess is stationary and isotropic and the size functional Σ is the kth intrinsic volume, fork ∈ 2, . . . , d. It turned out that zero cells which are large in this sense tend to becomespherical if the size functional tends to infinity. The case k = 1, where the size functionalis a multiple of the mean width and thus also of the parameter functional, was excluded.Generally, our method breaks down if the size functional Σ is proportional to the parameterfunctional Φ. For this choice of size functional, every convex body K ∈ Kd

o is an extremalbody, hence P(ϑ(Z0) ≥ ε) = 0 for every ε > 0. Theorem 1 holds trivially in this case, and doesnot distinguish between different shapes. In Section 7, we consider the special case where thedirectional distribution ϕ has finite support, and we prove that for the choice Σ = Φ a limitshape of the zero cell does, in fact, not exist. We are uncertain whether this example shouldlead one to speculate about a corresponding result for the uniform directional distribution. Itis interesting, in this connection, to recall a statement of Miles [10] in his heuristic approach.He considered stationary, isotropic Poisson line processes in the plane, where the mean widthis, up to a constant factor, the perimeter. Miles stated that the shape of typical cells withlarge perimeter tends to circular shape. It would be interesting to give a rigorous proof ofthis assertion (if true), to extend it to higher dimensions, and to see whether the analogue istrue for zero cells.

In the formulation of Theorem 1, we have preferred the condition Σ(Z0) ∈ [a, b), whichhas positive probability. However, our results are strong enough to allow also a treatmentof the conditional probability of the event ϑ(Z0) ≥ ε under the condition Σ(Z0) = a. Suchconditional probabilities appeared in the original formulation of Kendall’s problem. Underthe assumptions of Theorem 1, we will prove in Section 9 that

P(ϑ(Z0) ≥ ε | Σ(Z0) = a) ≤ c exp−c0f(ε)ar/kλ

(9)

for almost all a, with suitable positive constants c0, c.

In [3], Goldman described the asymptotic behaviour of the distribution function of thearea of the zero cell of an isotropic, stationary Poisson line process in the plane. We willextend Goldman’s result to the general situation to which Theorem 1 refers. While Goldman’smethod seems to be restricted to two dimensions, we can use some techniques developed forthe proof of Theorem 1 to obtain the generalization stated in Theorem 2. This theorem willbe proved in Section 8.

5

Theorem 2. If X and Σ are as above, then

lima→∞

a−r/k ln P(Σ(Z0) ≥ a) = −τλ.

Recall that the constant τ is given by

τ = minΦ(K) : K ∈ Kdo , Σ(K) = 1,

and equivalently by

e−τλ = maxP(K ⊂ Z0) : K ∈ Kdo , Σ(K) = 1.

3 Auxiliary Results

The expository section 2 contained a few unproved assertions; first we give the proofs ofthese. We begin with the inequality (4).

We note that the set K(1) := K ∈ Kdo : Φ(K) = 1 is compact. In fact, let K ∈ K(1),

let R be such that K ⊂ RBd and R is minimal. There exists x ∈ K with ‖x‖ = R, hence

1 = Φ(K) ≥ 1r

∫Sd−1

(〈x,u〉+)r ϕ(du) ≥ c(ϕ, r)Rr (10)

with a positive constant c(ϕ, r), since ϕ is not concentrated on a closed hemisphere. ThusK ⊂ (1/c(ϕ, r))1/rBd. Since Kd

o is closed and Φ is continuous, K(1) is compact.

On K(1), the continuous function Σ attains a maximum. It is positive, since K(1) containsa dilate of any body different from o in Kd

o , and the functional Σ is homogeneous and notidentically zero; let τ−k/r denote the value of the maximum. By homogeneity, we have

Φ(K) ≥ τ Σ(K)r/k for K ∈ Kdo . (11)

As defined in Section 2, a convex body K ∈ Kdo different from o for which equality holds

in (11) is called an extremal body.

Next we show that the minimum in (6) is attained. The subset of Kdo consisting of the

extremal bodies together with o is closed, hence it follows from the continuity propertiesof the metric that δ(K, ·) attains a minimum δ0 on this set. We have to show that theminimum is attained at an extremal body. Suppose the minimum is attained at o. ThenK ⊂ δ0B

d. By homogeneity, there is an extremal body M ⊂ δ0Bd. Since o ∈ K, we have

M ⊂ K + δ0Bd, and from o ∈ M we get K ⊂ M + δ0B

d. Thus, the minimum δ0 is alsoattained at the extremal body M .

To establish the existence of the strengthening (7), let ϑ be a deviation functional. We mayassume that there exist K ∈ Kd

o and ε0 > 0 with Σ(K) > 0 and ϑ(K) ≥ ε0, since otherwisethe assertion is trivial. By homogeneity, there is K ∈ Kd

o with Σ(K) > 0, ϑ(K) ≥ ε0, andΦ(K) = 1. Let 0 < ε ≤ ε0. On the nonempty compact set K ∈ Kd

o : Φ(K) = 1, ϑ(K) ≥ ε,the continuous function Σ attains a positive maximum, say at K0, which we write as τ

−k/rε .

We have τε > τ , since otherwise K0 would be an extremal body and, hence, ϑ(K0) = 0.Thus, we can write τε = (1 + f(ε))τ with f(ε) > 0. By homogeneity,

Φ(K) ≥ (1 + f(ε))τΣ(K)r/k if K ∈ Kdo and ϑ(K) ≥ ε. (12)

6

We put f(0) = 0 and f(ε) := f(ε0) for ε > ε0. The continuity of f is easy to check. Thus, fis a stability function for Φ,Σ, ϑ.

In the cases where the support of the directional distribution ϕ is not the whole unitsphere, the zero cells belong to a special class of convex bodies, which we now introduce.

Let Ω := suppϕ, the support of the directional distribution ϕ. This is a closed set onSd−1, not lying in a closed hemisphere. We say that a convex body K ∈ Kd is ϕ-adapted if

K =⋂

u∈Ω

H−(u, h(K, u)),

that is, if K is the intersection of those of its supporting halfspaces that have an outer unitnormal vector in the support of ϕ. The class of all ϕ-adapted convex bodies in Kd

o is denotedby Kϕ. For the inequality (4), there always exist extremal bodies belonging to Kϕ. In fact,let K be an extremal body. Then also

K ′ :=⋂

u∈Ω

H−(u, h(K, u))

is an extremal body. This follows from Φ(K ′) = Φ(K) (since the integrands in (3) for K ′

and K agree on the support of ϕ) and, by monotonicity, Σ(K ′) ≥ Σ(K) (which then impliesΣ(K ′) = Σ(K)). Clearly, the body K ′ is ϕ-adapted.

There may also exist extremal bodies which are not in Kϕ. In the context of asymptoticshapes of large cells, these bodies are irrelevant.

4 Limit Shapes and Asymptotic Shapes

Before proving Theorem 1, we want to demonstrate which implications this theorem has fora very general version of Kendall’s problem, that is, for the existence of limit shapes of largeCrofton cells, and more generally for a study of asymptotic shapes. Immediately from (8) weget

lima→∞

P(ϑ(Z0) < ε | Σ(Z0) ≥ a) = 1

for every ε > 0. Roughly, this shows that zero cells which are large in the sense of Σ have asmall deviation from an extremal body, with high probability.

In order to draw precise conclusions about the existence of limit shapes, we introducea suitable notion of shape. It is common to consider two convex bodies to be of the sameshape if they are similar to each other. We need a more general notion. Let G be a subgroupof the group S of similarities of Rd which contains the group D of all positive dilatations.Every such group will be called an admissible group. A typical example is the group H ofhomotheties (dilatations followed by translations). Two convex bodies K, M ∈ Kd have thesame G-shape, also written as K ∼G M , if K = gM with some g ∈ G. The quotient spaceSG := Kd/∼G is called the space of G-shapes. Let sG : Kd → SG be the canonical projection,thus sG(K) = gK : g ∈ G is the class of all convex bodies in Kd having the same G-shapeas K.

Let the Poisson hyperplane process X, the zero cell Z0 and the size functional Σ be asabove.

7

Definition. The conditional law of the G-shape of Z0, given the lower bound a for the sizeΣ, is the probability measure µa on SG defined by

µa := P(sG(Z0) ∈ · | Σ(Z0) ≥ a).

A shape sG(B), where B ∈ Kdo , is the limit shape of Z0 with respect to Σ if

lima→∞

µa = δsG(B) weakly,

where δsG(B) denotes the Dirac measure concentrated at sG(B).

Now we can formulate a general theorem on the existence of limit shapes.

Theorem 3. Let X, Z0,Σ be as above. Suppose there exists an admissible group G such thatthe extremal bodies of the inequality (4) have a unique G-shape sG(B). Then sG(B) is thelimit shape of Z0 with respect to Σ.

Proof. We deduce this from Theorem 1, assuming that all data are as given in that theoremand ϑ is chosen according to (5). For proving the asserted weak convergence of the measureµa, we have to show that

lim supa→∞

µa(C) ≤ δsG(B)(C) (13)

for every closed set C ⊂ SG. Let C be closed and not empty, without loss of generality.Relation (13) holds trivially if sG(B) ∈ C, hence we assume that sG(B) /∈ C. We set K1 :=K ∈ Kd

o ∩ s−1G (C) : Σ(K) = 1, then ϑ > 0 on K1. In fact, a body K ∈ K1 with ϑ(K) = 0

would be an extremal body and hence satisfy sG(K) = sG(B) /∈ C.

We show that ϑ attains a minimum on K1. Put ε := infϑ(K) : K ∈ K1. There isa sequence (Ki)i∈N in K1 such that ϑ(Ki) → ε for i → ∞. By the definition of ϑ and theestimate (10), this sequence is bounded and hence has a subsequence converging to someconvex body K0 ∈ Kd

o , satisfying sG(K0) ∈ C (since C is closed and sG is continuous) andΣ(K) = 1, thus K0 ∈ K1. It follows that ε > 0. For any K ∈ Kd

o ∩ s−1G (C) with Σ(K) > 0,

we can choose α > 0 with Σ(αK) = 1; then αK ∈ s−1G (C) (since G contains the positive

dilatations), and ϑ(αK) = ϑ(K). Thus ϑ ≥ ε on Kdo ∩ s−1

G (C) ∩ K : Σ(K) > 0 and, hence,Kd

o ∩ s−1G (C) ∩ K : Σ(K) > 0 ⊂ K ∈ Kd

o : ϑ(K) ≥ ε. This gives

µa(C) = P(Z0 ∈ Kdo ∩ s−1

G (C) | Σ(Z0) ≥ a) ≤ P(ϑ(Z0) ≥ ε | Σ(Z0) ≥ a) → 0

for a →∞, by Theorem 1, and thus proves Theorem 3.

In cases where limit shapes, as defined here, do not exist, one can still consider the ex-tremal bodies of (4) as asymptotic shapes for large zero cells, since (8) estimates the deviationof the D-shapes of large zero cells from the D-shapes of the extremal bodies.

Closer to the original formulation of Kendall’s problem, we can also consider the followingvariant.

Definition. The conditional law of the G-shape of Z0, given the value a for the size Σ, is theprobability measure νa on SG defined (for PΣ(Z0)-almost all a) by

νa := P(sG(Z0) ∈ · | Σ(Z0) = a).

8

A shape sG(B), where B ∈ Kdo , is the K-limit shape of Z0 with respect to Σ if

lima→∞

νa = δsG(B) weakly.

To justify this definition, we remark that Z0 is a random variable with values in themeasurable space (Kd,B(Kd)) (B denotes the Borel σ-algebra), and Σ(Z0) is a random vari-able with values in (R,B(R)). Since Kd is a Polish space, the regular conditional proba-bility distribution of Z0 with respect to Σ(Z0) exists, that is, there is a Markov kernel Qfrom (R,B(R)) to (Kd,B(Kd)) such that Q(·, B) is a version of the conditional probabilityP(Z0 ∈ B | Σ(Z0) = ·), for each B ∈ B(Kd). Thus, for each a ∈ R, the probability measureQ(a, ·) is the conditional distribution of Z0 under the hypothesis that Σ(Z0) = a; for eachB ∈ B(Kd) we have

Q(a,B) = P(Z0 ∈ B | Σ(Z0) = a) for PΣ(Z0)-almost all a.

It is now clear that νa as defined above is a probability measure, and that with the aidof inequality (9) we can obtain an analogue of Theorem 3 for K-limit shapes, with verballythe same proof.

5 Special Examples

Before proving Theorem 1 in the next section, we want to present some special examples,where the size of the zero cells is measured by functionals of geometric interest.

In the following, we denote by σ the normalized spherical Lebesgue measure on the sphereSd−1 and by κk the volume of the k-dimensional unit ball, thus κk = πk/2/Γ(1 + k/2).

(1) The zero cell of a stationary Poisson hyperplane mosaic; the size measured by the volume

This higher-dimensional version of Kendall’s problem, extended to the non-isotropic case,was treated in [4]. In this case, r = 1, Σ = Vd, and the parameter functional (with its presentnormalization) can be expressed as a mixed volume:

Φ(K) = dV1(B,K) = dV (B, . . . , B,K).

Here B is the convex body with centre o for which the directional distribution ϕ is the areameasure; it exists by Minkowski’s existence theorem. The isoperimetric inequality (4) is nowMinkowski’s classical inequality

V1(B,K)d ≥ Vd(B)d−1Vd(K).

Equality holds if and only if K is homothetic to B. Hence, sH(B) (the set of convex bodieshomothetic to B) is the limit shape of the zero cell with respect to the volume. If thehyperplane process is isotropic, then B is a ball, thus we get a higher dimensional version ofKendall’s original problem.

A deviation functional with a simple intuitive meaning is given by

rB(K) := infβ/α− 1 : αB ⊂ K + z ⊂ βB, z ∈ Rd, α, β > 0. (14)

9

A stability version of Minkowski’s inequality due to Groemer then leads to the followingdeviation estimate:

P(rB(Z0) ≥ ε | V (Z0) ∈ [a, b)) ≤ c exp−c0ε

d+1a1/dλ

.

(2) The typical cell of a stationary Poisson–Voronoı mosaic; the size measured by the kthintrinsic volume Vk, k ∈ 1, . . . , n

In this case, which was treated in [5], r = d, Σ = Vk, the directional distribution ϕ is therotation invariant probability measure σ, hence the parameter functional is given by

Φ(K) =1d

∫Sd−1

h(K, u)d σ(du).

The isoperimetric inequality (4) now reads

Φ(K) ≥ 1d

(κd−k(dk

)κd

)d/k

Vk(K)d/k. (15)

It is obtained by combining Holder’s inequality with the Aleksandrov-Fenchel inequalities.The extremal bodies are precisely the balls with centre o, hence the set of centred balls isthe limit shape of the typical cell Z with respect to Vk.

A convenient deviation functional is given by

ϑ(K) :=Ro(K)− ρo(K)Ro(K) + ρo(K)

, (16)

where Ro(K) (respectively ρo(K)) is the radius of the smallest (largest) ball with centre ocontaining K (contained in K). Using this deviation functional, a stability version of (15)can be proved, and the estimate

P(ϑ(Z) ≥ ε | Vk(Z) ∈ [a, b)) ≤ c exp−c0ε

(d+3)/2ad/kλ

is obtained.

Now we describe new cases, to which our extended general theorems can be applied. Incontrast to the previously established results, we can now have non-spherical limit shapeseven for zero cells of isotropic tessellations.

(3) The zero cell of a stationary, isotropic Poisson hyperplane mosaic; the size measured bythe diameter D

If K ∈ Kd, then K contains a segment of length D(K), without loss of generality with centreat o. We conclude that

Φ(K) =∫

Sd−1

h(K, u) σ(du) ≥ κd−1

dκdD(K),

10

with equality if and only if K is a segment. Thus, the limit shape of Z0 with respect to thediameter is the class of segments.

An intuitively natural deviation functional can be defined by

ϑ(K) := minα ≥ 0 : S ⊂ D(K)−1K ⊂ S + αBd, S a unit segment

for K ∈ Kd with dim K > 0.

Suppose that K ∈ Kd is a convex body of positive dimension with ϑ(K) ≥ ε > 0. Thereare a unit segment S ⊂ D(K)−1K and a point x ∈ K/D(K) on the boundary of S + εBd.The (possibly degenerate) triangle T := conv(S ∪ x) has perimeter L(T ) ≥ 1 +

√1 + 4ε2.

If w denotes the mean width in Rd, we have 2Φ = w and w(T ) = (κd−1/dκd)L(T ). This gives

Φ(D(K)−1K) ≥ Φ(T ) =12w(T ) ≥ κd−1

2dκd

(1 +

√1 + 4ε2

)and hence the stability estimate

Φ(K) ≥ κd−1

dκd(1 + ε2/2)D(K) if ϑ(K) ≥ ε,

for ε ≤ 1. Hence, we obtain the deviation estimate

P(ϑ(Z0) ≥ ε | D(Z0) ∈ [a, b)) ≤ c exp−c0ε

2aλ

.

(4) The typical cell of a stationary Poisson–Voronoı mosaic; the size measured by the largestdistance of a vertex from the nucleus

The largest distance of a vertex of Z0 from o is given by the centred circumradius Ro(Z0).If K ∈ Kd

o , then K contains a segment of length Ro(K) with one endpoint at o. This gives

Φ(K) =1d

∫Sd−1

h(K, u)d σ(du) ≥ 12d−1d

Rdo.

Equality holds if and only if K coincides with the chosen segment. Thus, in this case thelimit shape is the class of all segments with one endpoint at the origin.

(5) The zero cell of a stationary, isotropic Poisson hyperplane mosaic; the size measured bythe thickness θ

The thickness θ(K) of the convex body K is the smallest distance between any two parallelsupporting hyperplanes of K, thus θ(K) = minh(K, u) + h(K,−u) : u ∈ Sd−1. This gives

Φ(K) =12

∫Sd−1

[h(K, u) + h(K,−u)]σ(du) ≥ 12θ(K),

with equality if and only if K is a body of constant width. Hence, the asymptotic shape ofZ0 with respect to the thickness is the class of convex bodies of constant width.

(6) The zero cell of a stationary, nonisotropic Poisson hyperplane process; the size measuredby the inradius ρ

11

For a convex body K ∈ Kd, the inradius ρ(K) is the radius of a largest ball contained in K.For the zero cell Z0 of a stationary and isotropic Poisson hyperplane process X it was provedin [5, Theorem 3] that the limit shape with respect to the inradius is the class of balls. Weare now in a position to treat the nonisotropic case, where other limit shapes appear. In thiscase, the consideration of ϕ-adapted convex bodies is essential.

Since the process X is stationary, the directional distribution ϕ can be chosen as an evenmeasure, hence the parameter functional

Φ(K) =∫

Sd−1

h(K, u) ϕ(du), K ∈ Kdo ,

is translation invariant. We may therefore assume that o is the centre of a largest ballcontained in K. Then h(K, u) ≥ ρ(K), hence

Φ(K) ≥ ρ(K). (17)

Equality holds if and only if h(K, u) = ρ(K) for all u in the support of the measure ϕ. Thus,equality in (17) holds for the convex body

Bϕ :=⋂

u∈supp ϕ

H−(u, 1),

and for K ∈ Kϕ equality in (17) holds if and only if K is homothetic to Bϕ. We claimthat sH(Bϕ) is the limit shape of Z0 with respect to ρ. This cannot be deduced immediatelyfrom Theorem 3, since in general there are many extremal bodies not in Kϕ, and hence nothomothetic to Bϕ. Note, however, that the zero cell Z0 belongs to Kϕ with probability one.

To show that sH(Bϕ) is the limit shape of Z0 with respect to ρ, we modify the proof ofTheorem 3 (for G = H) in the following way. Let C ⊂ SH be a closed set with sH(Bϕ) /∈ C.Since Z0 ∈ Kϕ almost surely, we have

µa(C) = P(sH(Z0) ∈ C | ρ(Z0) ≥ a) = µa(C ∩ sH(Kϕ)).

In the proof of Theorem 3, we replace Kdo by Kϕ and Σ by ρ. The following must be

shown: if a sequence (Ki)i∈N in Kϕ converges to a convex body K0 with ρ(K0) = 1, thenK0 ∈ Kϕ. Assume this were false. Then

K0 6= K∗0 :=

⋂u∈supp ϕ

H−(u, h(K0,u)).

Since K0 ⊂ K∗0 , there is a point x ∈ intK∗

0 such that x /∈ K0 and hence a number α > 0 suchthat the ball Bd(x, α) with centre x and radius α satisfies Bd(x, α) ⊂ K∗

0 and Bd(x, α)∩K0 =∅. For all sufficiently large i we have |h(Ki,u)− h(K0,u| ≤ α for u ∈ suppϕ and thus

x ∈⋂

u∈supp ϕ

H−(u, h(Ki,u)) = Ki.

But this implies x ∈ K0, a contradiction. Thus, K0 ∈ Kϕ. Now the proof can be completedas that of Theorem 3.

We remark that for the preceding proof it was essential that K0 has nonempty interior.Otherwise, we cannot conclude from Ki ∈ Kϕ and Ki → K0 that K0 ∈ Kϕ. A counterexampleis provided by [1, Example 20.6].

12

A stability improvement of (17) involving a simple geometrically reasonable deviationfunctional, like (14) or (16), can apparently not be achieved without special assumptions onthe directional distribution ϕ.

(7) The zero cell of a nonstationary, nonisotropic Poisson hyperplane process; the size mea-sured by the centred inradius ρ0

In the nonstationary case, the parameter functional

Φ(K) =1r

∫Sd−1

h(K, u)r ϕ(du)

is not translation invariant, therefore we replace the inradius ρ(K) by the centred inradiusρo(K). As above, we obtain

Φ(K) ≥ 1rρo(K)r for K ∈ Kd

o ,

with equality for K ∈ Kϕ if and only if K is a dilate of Bϕ. Similarly as before, it can beshown that sD(Bϕ) is the limit shape of Z0 with respect to the centred inradius ρo.

Up to now, we have only considered size functionals which are isotropic, that is, invariantunder rotations. This is not required by our general result. Extending a study of Miles [10,Section 6] in the plane, we now measure the size by the width in a given direction.

(8) The zero cell of a stationary, isotropic Poisson hyperplane mosaic; the size measured bythe width wv in a given direction v

The width wv(K) of the convex body K in the given direction v ∈ Sd−1 is the distancebetween the two supporting hyperplanes of K orthogonal to v. Since K contains a segmentof length at least wv(K), one obtains, similarly as in Example (3) above, the estimate

Φ(K) ≥ κd−1

dκdwv(K),

with equality if and only if K is a segment parallel to v. Hence, the limit shape of the zerocell with respect to wv is the class of segments of direction v.

Many more cases could be considered, but often they lead to unsolved geometric problemsabout the determination of the extremal bodies of the isoperimetric inequality (4). Forexample, the question for the asymptotic shape of the zero cell of a stationary, non-isotropicPoisson hyperplane process with respect to the kth intrinsic volume Vk leads to the problem ofmaximizing Vk(K) under the condition

∫Sd−1 h(K, u) ϕ(du) = 1, for a given even probability

measure ϕ not concentrated on a great subsphere. Another example is the choice of thecircumradius as a size functional. In the stationary and isotropic case, one would have todetermine the convex bodies with given circumradius and minimal mean width. It is plausiblethat these are the segments, and for d = 2 this is easy to see, but in higher dimensions aproof seems lacking.

13

6 Proof of Theorem 1

As an introduction to the proof of Theorem 1, we extend a heuristic argument from [4], tryingto make plausible why an estimate as in Theorem 1 can be expected. In these heuristics,we restrict ourselves to an interval [a,∞), with a > 0. We have to estimate the conditionalprobability

P(ϑ(Z0) ≥ ε | Σ(Z0) ≥ a) =P(ϑ(Z0) ≥ ε, Σ(Z0) ≥ a)

P(Σ(Z0) ≥ a).

Estimation of the denominator is easy. There exists an extremal body B ∈ Kdo . Let Ba be

the dilate of B with Σ(Ba) = a. Then, by the monotonicity of Σ,

P(Σ(Z0) ≥ a) ≥ P(X(HBa) = 0) = exp−Φ(Ba)λ.

Since Ba is an extremal body, we have

Φ(Ba) = τΣ(Ba)r/k = τar/k, (18)

henceP(Σ(Z0) ≥ a) ≥ exp−τar/kλ. (19)

For the estimation of the numerator, we try to compare the occurring zero cells witha deterministic convex body with similar properties, that is, not cut by hyperplanes of theprocess, with large size and large deviation from B. Suppose that K ∈ Kd

o is a convex bodysatisfying

ϑ(K) ≥ ε > 0 and Σ(K) ≥ a.

Then, by (7),

P(X(HK) = 0) = exp−Φ(K)λ ≤ exp−(1 + f(ε))τar/kλ.

Heuristically, we hope that here we may replace the deterministic convex body K satisfying

X(HK) = 0, ϑ(K) ≥ ε, Σ(K) ≥ a

by the random polytope Z0 satisfying

X(HαZ0) = 0 ∀α ∈ (0, 1), ϑ(Z0) ≥ ε, Σ(Z0) ≥ a,

at the cost of only a slight weakening of the inequality, say

P(ϑ(Z0) ≥ ε, Σ(Z0) ≥ a) ≤ c′ exp−(1 + c′′f(ε))τar/kλ (20)

with c′, c′′ > 0. If (20) can be proved, then together with (19) this implies

P(ϑ(Z0) ≥ ε | Σ(Z0) ≥ a) ≤ c′ exp−c′′f(ε)τar/kλ,

which is of the form asserted in Theorem 1.

The bulk of the proof will consist in replacing this heuristic argument by precise reasoning.The proof is an extension of the one given in [4] for the case of a stationary Poisson hyperplaneprocess, the volume functional, and a special deviation functional. We merely quote thoseparts of the proof in [4] which require only obvious changes, but we give full proofs for theparts which require extended arguments or where simplifications have been possible.

14

To simplify the notation, we write

HnK := HK × · · · × HK (n times),

µn := µ⊗ · · · ⊗ µ (n times),

and for given hyperplanes H1, . . . ,Hn ∈ Hd, we write

H−1 ∩ · · · ∩H−

n =: P (H(n)).

Convention about constants. In the following, c1, c2, . . . denote positive constants whichmay depend on the dimension d, the directional distribution ϕ, the distance exponent r, andthe size functional Σ. If a constant depends on additional data, these are either indicated asarguments or mentioned in the text.

For the proof of Theorem 1, we suppose that all data are given as assumed in that theoremand as explained in Section 2.

In contrast to the special case just considered heuristically, we will have to admit generalintervals (a, b) = a(1, 1 + h) as ranges for Σ(Z0). In a first stage, this is only possible forsufficiently small positive h.

The proof of Theorem 1 is preceded by a number of lemmas. Our first lemma is a counter-part to inequality (19) for certain bounded intervals. Its proof extends that of Lemma 2 in [5](rather than that of Lemma 3.2 in [4]), but must be changed for the present situation, whereextremal bodies may be of lower dimension and the support of the directional distribution ϕneed not be all of Sd−1.

Lemma 1. For each β > 0, there are constants h0 > 0, N ∈ N and c > 0, depending onlyon ϕ, r, Σ and β, such that for a > 0 and 0 < h < h0,

P(Σ(Z0) ∈ a(1, 1 + h)) ≥ c h(ar/kλ)N exp−(1 + β)τar/kλ.

Proof. As shown in Section 3, there exist extremal bodies in the class Kϕ of ϕ-adapted bodies.We choose such a body B with Σ(B) = 1, then Φ(B) = τ . For given data Φ and Σ, we makea definite choice of B, so that constants depending on B depend, in fact, on Φ and Σ.

For given numbers a > 0 and γ > 0, we set

B(a, γ) := a1/k(B + γBd).

We choose γ so small that

Φ(B(a, γ)) = τar/k Φ(B + γBd)Φ(B)

≤ (1 + β)τar/k, (21)

which is possible by the continuity of Φ. The choice of γ depends on Φ, B and β.

Further, we choose a number h0 > 0, depending only on B and γ, so that

(1 + h0)1/k[h(B, u) + γ/2] ≤ h(B, u) + γ for u ∈ Sd−1. (22)

In the following, we assume that 0 < h < h0.

15

For a convex body K ∈ Kdo we set

H∂K := H ∈ Hd : H touches K.

By definition, H touches K if H meets K but not the interior of K. Thus, H ∈ H∂K if andonly if h(K, u) > 0 and H = H(u, h(K, u)) for some u ∈ Sd−1. We define

Mn :=

(H1, . . . ,Hn) ∈ Hn−1B(1,γ/2) ×H∂B : P (H(n)) ⊂ B(1, γ/2), Σ(P (H(n))) ≥ 1

.

Let N be an arbitrary positive integer (it will be specified later). The crucial eventΣ(Z0) ∈ a(1, 1 + h) certainly occurs if the body B(a, γ) is hit by precisely N hyperplanesH1, . . . ,HN of X and the polytope P (H(N)) is contained in B(a, γ) and satisfies Σ(P (H(N))) ∈a(1, 1 + h). By the Poisson property, under the condition that X(HB(a,γ)) (= card(X ∩HB(a,γ))) = N , the process X ∩ HB(a,γ) is stochastically equivalent to the process definedby the set of N independent, identically distributed random hyperplanes with distributionΘ(· ∩ HB(a,γ))/Θ(HB(a,γ)). Recall that Θ = λµ and µ(HB(a,γ)) = Φ(B(a, γ)). Thus we get

P(Σ(Z0) ∈ a(1, 1 + h))

≥ P(X(HB(a,γ)) = N) P(P (X ∩HB(a,γ)) ⊂ B(a, γ),

Σ(P (X ∩HB(a,γ))) ∈ a(1, 1 + h) | X(HB(a,γ)) = N)

=λN

N !exp−Φ(B(a, γ))λ

∫HN

B(a,γ)

1P (H(N)) ⊂ B(a, γ)

×1Σ(P (H(N))) ∈ a(1, 1 + h)µN (d(H1, . . . ,HN )). (23)

Let H ∈ Hd. If the translate of H through o does not contain B, we denote by b(H) theunique positive number for which b(H)H ∈ H∂B; otherwise, we set b(H) := 0. Assume thatH1, . . . ,HN ∈ Hd satisfy the conditions

(i) b(HN )(H1, . . . ,HN ) ∈MN ,

(ii) Σ(P (H(N))) ∈ a(1, 1 + h).

We use the standard representations Hi = H(vi, ti), i = 1, . . . , N . Then, in particular,

b(HN ) =h(B, vN )

tN

if b(HN ) > 0 (note that b(HN ) > 0 is implied by (i)). Using (i), (ii) and the definition ofMN , we get

1 ≤ Σ(b(HN )P (H(N))) = b(HN )kΣ(P (H(N))) ≤(

h(B, vN )tN

)k

a(1 + h),

thustN

h(B, vN )≤ a1/k(1 + h)1/k. (24)

Using (i) and the definition of MN , we find for i = 1, . . . , N that

tih(B, vN )tN

≤ h(B(1, γ/2),vi);

16

hence (24) and (22) give

ti ≤ a1/k(1 + h)1/k[h(B, vi) + γ/2] ≤ h(B(a, γ),vi),

thusHi ∈ HB(a,γ) for 1 = 1, . . . , N.

Finally, (i) and the definition of MN together with (24) and (22) imply that

P (H(N)) ⊂tN

h(B, vN )B(1, γ/2) ⊂ a1/k(1 + h)1/kB(1, γ/2) ⊂ B(a, γ).

Thus, if H1, . . . ,HN satisfy (i), (ii), then the indicator functions in the integral (23) are equalto one. We deduce that

P(Σ(Z0) ∈ a(1, 1 + h)) ≥ λN

N !exp−Φ(B(a, γ))λ · I, (25)

where

I :=∫HN

1b(HN )(H1, . . . ,HN ) ∈MN1Σ(P (H(N))) ∈ a(1, 1 + h)µN (d(H1, . . . ,HN )).

With an arbitrary number η > 0, we can estimate

I ≥∫

Sd−1

. . .

∫Sd−1

∫ ∞

0. . .

∫ ∞

01 (H(v1, t1), . . . ,H(vN , tN )) ∈ (tN/h(B, vN ))MN

×1Σ(H−(v1, t1) ∩ · · · ∩H−(vN , tN )

)∈ a(1, 1 + h)

1h(B, vN ) ≥ η

×(t1 · · · tN )r−1 dt1 . . .dtN ϕ(dv1) . . . ϕ(dvN ).

In the inner integrals, we introduce new variables t1, . . . , tN−1, z by ti = zti for i = 1, . . . , N−1and tN = zh(B, vN ); then we first carry out the integration with respect to z. This gives

I ≥∫

Sd−1

. . .

∫Sd−1

∫ ∞

0. . .

∫ ∞

01h(B, vN ) ≥ η

×1 (H(v1, t1), . . . ,H(vN−1, tN−1),H(vN , h(B, vN ))) ∈MN

×1

zkΣ(H−(v1, t1) ∩ · · · ∩H−(vN−1, tN−1) ∩H−(vN , h(B, vN ))

)∈ a(1, 1 + h)

×zrN−1h(B, vN )r(t1 · · · tN−1)r−1 dt1 . . .dtN−1 dz ϕ(dv1) . . . ϕ(dvN )

=∫

Sd−1

. . .

∫Sd−1

∫ ∞

0. . .

∫ ∞

01h(B, vN ) ≥ η

×1 (H(v1, t1), . . . ,H(vN−1, tN−1),H(vN , h(B, vN ))) ∈MN

×arN/kh(B, vN )r

rNΣ(H−(v1, t1) ∩ · · · ∩H−(vN−1, tN−1) ∩H−(vN , h(B, vN ))

)−rN/k

×[(1 + h)rN/k − 1

](t1 · · · tN−1)r−1 dt1 . . .dtN−1 ϕ(dv1) . . . ϕ(dvN )

≥ arN/kηr

rNΣ(B(1, γ/2))−rN/k rN

kh · J,

17

where

J :=∫HN−1

∫Sd−1

1(H1, . . . ,HN−1,H(u, h(B, u))) ∈MN1h(B, u) ≥ η

ϕ(du) µN−1(d(H1, . . . ,HN−1)).

So far, N and η were arbitrary. We must now show that there are numbers N ∈ N andη > 0 such that

J ≥ c1(β) > 0.

For this, we choose a vector u0 ∈ suppϕ with

h(B, u0) > 0

(it exists since B 6= o) and a countable dense subset u1,u2, . . . of supp ϕ. Then

B =∞⋂i=1

H−(ui, h(B, ui)). (26)

In fact, denote the right-hand side of (26) by B′, then B ⊂ B′ trivially. Let x ∈ Rd \B. SinceB ∈ Kϕ, there is a vector u ∈ suppϕ such that x /∈ H−(u, h(B, u)). A vector ui sufficientlyclose to u satisfies x /∈ H−(ui, h(B, ui)), hence x /∈ B′. This proves (26).

For n ∈ N, let

Pn :=n⋂

i=1

H−(ui, h(B, ui)),

then Pn ↓ B in the Hausdorff metric, as n → ∞ ([12], Lemma 1.8.1). Hence, we can choosea number N for which P ′N := PN−1 ∩H−(u0, h(B, u0)) ⊂ B(1, γ/4). For given data Φ andΣ and accordingly chosen B, we make a definite choice of such a polytope P ′N satisfyingP ′N ⊂ B(1, γ/4); it then depends only on Φ, Σ and β.

By continuity, we can choose a neighbourhood UN of u0 and a number η > 0 suchthat h(B, vN ) ≥ η for vN ∈ UN . Further, there exist neighbourhoods Ui ⊂ Sd−1 of ui,i = 1, . . . , N − 1, and a number α > 0 (depending on B, P ′N and γ) such that

vi ∈ Ui for i = 1, . . . , N and 0 ≤ ti ≤ α for i = 1, . . . , N − 1

impliesH(vi, h(B, vi) + ti) ∩B(1, γ/2) 6= ∅ for i = 1, . . . , N − 1

and

P :=N−1⋂i=1

H−(vi, h(B, vi) + ti) ∩H−(vN , h(B, vN )) ⊂ B(1, γ/2).

Since B ⊂ P , we have Σ(P ) ≥ Σ(B) = 1. This shows that

(H(v1, h(B, v1) + t1), . . . ,H(vN−1, h(B, vN−1) + tN−1),H(vN , h(B, vN ))) ∈MN

and hence that

J ≥(

1rαr

)N−1 N∏i=1

ϕ(Ui) =: c1(β) > 0. (27)

Here ϕ(Ui) > 0 follows from uj ∈ suppϕ for j = 0, . . . , N − 1.

18

Introducing (27) in the estimate for I and combining this with (25) and (21), we obtainthe assertion of Lemma 1.

In the following, D denotes the diameter. On the compact set K ∈ Kdo : D(K) = 1, the

continuous function Σ attains a (positive) maximum 1/ck2. By homogeneity, it follows that

D(K) ≥ c2Σ(K)1/k for K ∈ Kdo . For K ∈ Kd

o with Σ(K) > 0 we introduce the relativediameter ∆ by

∆(K) := D(K)/c2Σ(K)1/k.

For a > 0, ε ≥ 0, h > 0 and m ∈ N we define

Ka,ε,h(m) := K ∈ Kϕ : Σ(K) ∈ a(1, 1 + h), ϑ(K) ≥ ε, ∆(K) ∈ [m,m + 1)

andqa,ε,h(m) := P(Z0 ∈ Ka,ε,h(m)).

(Note that the condition ϑ(K) ≥ ε is trivially satisfied for ε = 0.) We have

∞∑m=1

qa,ε,h(m) = P(Σ(Z0) ∈ a(1, 1 + h), ϑ(Z0) ≥ ε). (28)

For the latter probability, Lemma 9 will provide the upper estimate which is crucial for theproof of Theorem 1. The reason for introducing the additional restriction ∆(Z0) ∈ [m,m+1)lies in the fact that it will allow us, thanks to Lemma 2, to consider in a first step only thezero cells lying in some fixed bounded set C. The delicate estimate of Lemma 5 obtained inthis way will later be used for small numbers m, while for large m the estimate of Lemma 3is sufficient. For technical reasons, we first consider only ranges for Σ(Z0) of the type a(1, 2).Lemmas 6, 7, 8 are needed to extend this to intervals a(1, 1+h), with sufficiently small h > 0.The completion of the proof then settles the case of general intervals.

By Pdo we denote the set of polytopes in Kd

o .

Lemma 2. For each m ∈ N:

(a) K ∈ Ka,0,1(m) implies K ⊂ c3ma1/kBd =: C,

(b) there exists a measurable map

Ka,0,1(m) ∩ Pdo 3 P 7→ v(P )

such that v(P ) is a vertex of P with ‖v(P )‖ ≥ c4ma1/k.

Proof. Let K ∈ Ka,0,1(m). Then

D(K) < (m + 1)c2Σ(K)1/k ≤ (m + 1)c2(2a)1/k ≤ c3ma1/k.

Since o ∈ K, this proves (a).

Further,D(K) ≥ c2mΣ(K)1/k ≥ c2ma1/k.

Hence, if P is a polytope in Ka,0,1(m), it has a vertex v such that ‖v‖ ≥ (1/2)c2ma1/k. Theexistence of the measurable map follows as in [4], proof of Lemma 4.3.

19

Lemma 3. For a > 0 and m ∈ N,

qa,0,1(m) ≤ c7 exp−c5mrar/kλ.

Proof. Let m ∈ N be given. Let C be the ball defined in Lemma 2. We will repeatedly makeuse of

qa,ε,1(m) =∞∑

N=d+1

P(X(HC) = N)P(Z0 ∈ Ka,ε,1(m) | X(HC) = N), (29)

where

pN := P(Z0 ∈ Ka,ε,1(m) | X(HC) = N)

=1

Φ(C)N

∫HN

C

1P (H(N)) ∈ Ka,ε,1(m)µN (d(H1, . . . ,HN )), (30)

since X is a Poisson process. In the present proof, ε = 0. Suppose that H1, . . . ,HN ∈ HC aresuch that P := P (H(N)) ∈ Ka,0,1(m). Let v(P ) be the vertex according to Lemma 2. Thisvertex is the intersection of d facets of P . Hence, there exists an index set J ⊂ 1, . . . , Nwith d elements such that

v(P ) =⋂i∈J

Hi.

We denote the segment [o,v(P )] by S(Hi, i ∈ J). The segment S = S(Hi, i ∈ J) satisfies

Hi ∩ relint S = ∅ for i ∈ 1, . . . , N \ J,

where relint denotes the relative interior. Since S ⊂ C, we have∫HC

1H ∩ S = ∅µ(dH) = Φ(C)− Φ(S).

We denote by U0 a closed segment of unit length and with one endpoint at o for which

Φ(U0) = minΦ([o,u]) : u ∈ Sd−1.

ThenΦ(C)− Φ(S) ≤ Φ(C)− |S|rΦ(U0),

where |S| is the length of S. Since ϕ is not concentrated on a closed hemisphere, we haveΦ(U0) > 0. Thus, we obtain

pN ≤(

N

d

)1

Φ(C)N

∫Hd

C

∫HN−d

C

1|S(Hj , j ∈ 1, . . . , d)| ≥ c4ma1/k

×1 S(Hj , j ∈ 1, . . . , d) ∩Hi = ∅ for i = d + 1, . . . , N

µN−d(d(Hd+1, . . . ,HN ))µd(d(H1, . . . ,Hd))

≤(

N

d

)Φ(C)−N

∫Hd

C

[Φ(C)− (c4ma1/k)rΦ(U0)]N−d µd(d(H1, . . . ,Hd))

=(

N

d

)Φ(C)d−N

[Φ(C)− 2c5m

rar/k]N−d

,

20

with c5 > 0. This leads to the estimate

qa,0,1(m) ≤∞∑

N=d+1

[Φ(C)λ]N

N !exp−Φ(C)λ

(N

d

)Φ(C)d−N

[Φ(C)− 2c5m

rar/k]N−d

=1d!

[Φ(C)λ]d exp−Φ(C)λ∞∑

N=d+1

1(N − d)!

[Φ(C)λ− 2c5m

rar/kλ]N−d

≤ 1d!

[Φ(C)λ]d exp−2c5m

rar/kλ

≤ c6(mrar/kλ)d exp−2c5m

rar/kλ

≤ c7 exp−c5m

rar/kλ

.

For a polytope P , we denote by ext P the set of vertices and by f0(P ) the number of verticesof P . The following approximation result will allow us to essentially restrict ourselves tozero cells with a bounded number of vertices. The proof of the lemma is the same as that ofLemma 5 in [5], with the obvious changes.

Lemma 4. Let α > 0 be given. There is a number ν ∈ N depending only on d, ϕ, r and αsuch that the following is true. For P ∈ Pd

o there exists a polytope Q = Q(P ) ∈ Pdo satisfying

ext Q ⊂ ext P , f0(Q) ≤ ν, and Φ(Q) ≥ (1 − α)Φ(P ). Moreover, there exists a measurableselection P 7→ Q(P ).

From now on we assume that, for given Φ, Σ and ϑ, a stability function f according to (7)has been chosen.

Lemma 5. For a > 0, m ∈ N and ε > 0,

qa,ε,1(m) ≤ c10(f, ε)mrdν exp−(1 + f(ε)/3)τar/kλ

,

where ν depends only on d, ϕ, r and ε.

Proof. Let B be the extremal body chosen in the proof of Lemma 1, and let Ba be the dilateof B with Σ(Ba) = a. For given m ∈ N, we define C as in Lemma 2 and use (29) and (30).Suppose that H1, . . . ,HN ∈ HC are such that P (H(N)) ∈ Ka,ε,1(m). By (7) and (18),

Φ(P (H(N))) ≥ (1 + f(ε))τΣ(P (H(N)))r/k ≥ (1 + f(ε))τar/k

= (1 + f(ε))Φ(Ba). (31)

Let α := f(ε)/(2 + f(ε)), then (1− α)(1 + f(ε)) = 1 + α.

We generalize the proof of Lemma 5.2 in [4]. By Lemma 4, there are ν = ν(d, ϕ, r, ε)vertices of P (H(N)) such that the convex hull Q(P (H(N))) =: Q(H(N)) =: Q of these verticessatisfies

Φ(Q) ≥ (1− α)Φ(P (H(N))). (32)

21

The inequalities (31) and (32) imply that

Φ(Q) ≥ (1 + α)Φ(Ba).

For each N -tuple (H1, . . . ,HN ) such that P (H(N)) ∈ Ka,ε,1(m), we make a definite choice ofQ = Q(H(N)), in such a way that Q(H(N)) is a measurable function of (H1, . . . ,HN ).

Excluding a set of N -tuples (H1, . . . ,HN ) of µN measure zero, we can assume that eachof the vertices of Q lies in precisely d of the hyperplanes H1, . . . ,HN , and the remaininghyperplanes are disjoint from Q. Hence, at most dν of the hyperplanes H1, . . . ,HN meetQ; let j ∈ d + 1, . . . , dν denote their precise number. Suppose that H1, . . . ,Hj are thehyperplanes meeting Q. Then there are subsets J1, . . . , Jf0(Q) ⊂ 1, . . . , j, each of cardinalityd, such that the intersections ⋂

i∈Jr

Hi, r = 1, . . . , f0(Q) ≤ ν,

yield the vertices of Q. In the following, the sum∑

(J1,...,Jν) extends over all ν-tuples ofd-element subsets of 1, . . . , j. To estimate the inner ((N − j)-fold) integral below, we makeuse of the fact that for any convex body K ⊂ C,∫

HC

1H ∩K = ∅µ(dH) = Φ(C)− Φ(K).

In this way, we obtain

P(Z0 ∈ Ka,ε,1(m) | X(HC) = N)Φ(C)N

≤dν∑

j=d+1

(N

j

)∫HN

C

1P (H(N)) ∈ Ka,ε,1(m)

1Hi ∩Q(H(N)) 6= ∅ for i = 1, . . . , j

×1Hi ∩Q(H(N)) = ∅ for i = j + 1, . . . , NµN (d(H1, . . . ,HN ))

≤dν∑

j=d+1

(N

j

) ∑(J1,...,Jν)

∫Hj

C

∫HN−j

C

1

Φ

(conv

ν⋃r=1

⋂i∈Jr

Hi

)≥ (1 + α)Φ(Ba)

×1

Hs ∩ conv

ν⋃r=1

⋂i∈Jr

Hi = ∅ for s = j + 1, . . . , N

µN−j(d(Hj+1, . . . ,HN ))µj(d(H1, . . . ,Hj))

≤dν∑

j=d+1

(N

j

)(j

d

)ν

[Φ(C)− (1 + α)Φ(Ba)]N−jΦ(C)j .

Summation over N gives

qa,ε,1(m) ≤∞∑

N=d+1

[Φ(C)λ]N

N !exp−Φ(C)λ

dν∑j=d+1

(N

j

)(j

d

)ν [Φ(C)− (1 + α)Φ(Ba)]N−j

Φ(C)N−j

=dν∑

j=d+1

(j

d

)ν [Φ(C)λ]j

j!exp−Φ(C)λ

∞∑N=j

1(N − j)!

[Φ(C)λ− (1 + α)Φ(Ba)λ]N−j

=dν∑

j=d+1

(j

d

)ν [Φ(C)λ]j

j!exp−(1 + α)Φ(Ba)λ.

22

Here Φ(Ba) = τar/k by (18), and by Lemma 2,

Φ(C) = Φ(c3ma1/kBd) = c8mrar/k.

Thus we get

qa,ε,1(m) ≤ c9(ε)[(ar/kλ)dν + 1

]mrdν exp

−(1 + α)τar/kλ

≤ c10(ε)mrdν exp

−(1 + f(ε)/3)τar/kλ

,

since f(ε) < 1.

The next steps serve to extend the estimate for qa,ε,1(m) to one for qa,ε,h(m), for h > 0.

Lemma 6. Let w > 0, h ∈ (0, 1/2), r ≥ 1 and p ≥ r − 1. Then∫ k√1+h

1spexp−wsrds ≤ b1hw[1 + (expb2w − 1)−1]

∫ k√2

1spexp−wsrds

with positive constants b1, b2 depending only on r and k.

Proof. Substituting s = x1/r and applying in turn the mean value theorems of integralcalculus and differential calculus, we get∫ k√1+h

1spexp−wsrds ≤ b1(r, k)h

1rη(p+1−r)/r exp−wη

with a suitable number η ∈ (0, (1 + h)r/k). As in the proof of Lemma 6.2 in [4], we obtain∫ k√2

1spexp−wsrds ≥ 1

rη(p+1−r)/r exp−wη 1

w[1− exp−b2(r, k)w] .

The assertion follows by combining these two inequalities.

The next lemma is stated in a general version, since different specializations are needed. Herefd−1(P ) denotes the number of facets of a polytope P .

Lemma 7. For n ∈ N, n ≥ d + 1 and a Borel set B ⊂ Kdo, let

R(B, n) := (H1, . . . ,Hn) ∈ (Hd)n : P (H(n)) ∈ B, fd−1(P (H(n))) = n.

Then

P(Z0 ∈ B, fd−1(Z0) = n) =λn

n!

∫R(B,n)

exp−Φ(P (H(n)))λ

µn(d(H1, . . . ,Hn)).

Proof. Let C ⊂ Rd be a ball with centre o and put

R(B, n, C) := (H1, . . . ,Hn) ∈ R(B, n) : P (H(n)) ⊂ C.

23

Then, for N ≥ n,

P(Z0 ∈ B, fd−1(Z0) = n, Z0 ⊂ C | X(HC) = N)

=1

Φ(C)N

∫HN

C

1P (H(N)) ∈ B, fd−1(P (H(N))) = n, P (H(N)) ⊂ C

µN (d(H1, . . . ,HN ))

=

(Nn

)Φ(C)N

∫R(B,n,C)

[Φ(C)− Φ(P (H(n)))

]N−nµn(d(H1, . . . ,Hn))

and hence

P(Z0 ∈ B, fd−1(Z0) = n, Z0 ⊂ C)

=∞∑

N=n

P(Z0 ∈ B, fd−1(Z0) = n, Z0 ⊂ C | X(HC) = N)[Φ(C)λ]N

N !exp −Φ(C)λ

= exp −Φ(C)λ

× λn

n!

∫R(B,n,C)

∞∑N=n

1(N − n)!

[Φ(C)λ− Φ(P (H(n)))λ

]N−nµn(d(H1, . . . ,Hn))

=λn

n!

∫R(B,n,C)

exp−Φ(P (H(n)))λ

µn(d(H1, . . . ,Hn)).

We apply this to an increasing sequence of balls covering Rd and then obtain the assertionfrom the monotone convergence theorem.

Now let a > 0, ε ≥ 0, h > 0 and m ∈ N. We set

qa,ε,h(m, n) := P (Z0 ∈ Ka,ε,h(m), fd−1(Z0) = n) (33)

for n ∈ N, n ≥ d + 1. Next, we define

Rε(m,n) :=

(H1, . . . ,Hn) ∈ (Hd)n : ϑ(P (H(n))) ≥ ε, ∆(P (H(n))) ∈ [m,m + 1),

fd−1(P (H(n))) = n

andRa,ε,h(m,n) :=

(H1, . . . ,Hn) ∈ Rε(m,n) : Σ(P (H(n))) ∈ a(1, 1 + h)

.

In the following two lemmas, we introduce an additional parameter σ0. This can be putequal to 1 in the proofs of Theorems 1 and 2; it will only be needed for (41), which is appliedin Section 9.

Lemma 8. For m ∈ N, h ∈ (0, 1/2), ε ≥ 0 and ar/kλ ≥ σ0, where σ0 > 0 is a constant,

qa,ε,h(m) ≤ c13(σ0)h ar/kλ mrqa,ε,1(m).

Proof. We apply Lemma 7 with

B := K ∈ Ka,ε,h(m) : K ∈ Pdo , fd−1(K) = n.

24

Using standard representations Hi = H(ui, ti), we write the result as

qa,ε,h(m,n)

=λn

n!

∫Sd−1

. . .

∫Sd−1

∫ ∞

0. . .

∫ ∞

01 (H(u1, t1), . . . ,H(un, tn)) ∈ Ra,ε,h(m,n)

× exp−Φ

(H−(u1, t1) ∩ · · · ∩H−(un, tn)

)λ

× (t1 · · · tn)r−1 dt1 . . .dtn ϕ(du1) . . . ϕ(dun).

In the inner integrals, we introduce new variables t1, . . . , tn−1, z by ti = zti for i = 1, . . . , n−1and tn = z; then we first carry out the integration with respect to z. We exploit the factthat Rε(m,n) is closed under dilatations. Writing H(ui, ti) = Hi again for i = 1, . . . , n − 1,we obtain

qa,ε,h(m,n) =λn

n!

∫(Hd)n−1

∫Sd−1

1 (H1, . . . ,Hn−1,H(u, 1)) ∈ Rε(m,n)

×∫ ∞

01

zkΣ(H−

1 ∩ · · · ∩H−n−1 ∩H−(u, 1)

)∈ a(1, 1 + h)

× exp

−zrΦ

(H−

1 ∩ · · · ∩H−n−1 ∩H−(u, 1)

)λ

zrn−1 dz

×ϕ(du) µn−1(d(H1, . . . ,Hn−1)).

For the computation and estimation of the inner integral∫∞0 (· · · ) dz, we fix H1, . . . ,Hn−1,u

with (H1, . . . ,Hn−1,H(u, 1)) ∈ Rε(m,n), write

H−1 ∩ · · · ∩H−

n−1 ∩H−(u, 1) =: P,

and define za = za(H1, . . . ,Hn−1,u) by Σ(zaP ) = a. Then Σ(zak√

1 + hP ) = a(1 + h). SincezaP ∈ Ka,0,1(m), it follows from Lemma 2 that

c4ma1/k[o,v] ⊂ zaP ⊂ c3ma1/kBd,

where v is a suitable unit vector, hence

c11mrar/k ≤ Φ(zaP ) ≤ c12m

rar/k. (34)

In the subsequent computation, we substitute z = zas, then we apply Lemma 6, the inequal-ities (34), and reverse the substitution; this gives∫ ∞

0(· · · ) dz =

∫ zak√1+h

za

exp−Φ(zP )λzrn−1 dz

= zrna

∫ k√1+h

1exp−srΦ(zaP )λsrn−1 ds

≤ b1hΦ(zaP )λ[1 + (exp b2Φ(zaP )λ − 1)−1

]× zrn

a

∫ k√2

1exp−srΦ(zaP )λsrn−1 ds

≤ c13(σ0)har/kλmr

∫ zak√2

za

exp−Φ(zP )λzrn−1 dz.

25

Inserting this in the multiple integral representing qa,ε,h(m, n), we obtain the assertion of thelemma after a summation over n ∈ N, n ≥ d + 1.

Our last lemma is the counterpart to Lemma 1. Recall that the stability function f waschosen before the formulation of Lemma 5.

Lemma 9. Let ε > 0, h ∈ (0, 1/2) and ar/kλ ≥ σ0, where σ0 > 0 is a constant. Then

P(Σ(Z0) ∈ a(1, 1 + h), ϑ(Z0) ≥ ε) ≤ c15(f, ε, σ0)h exp−(1 + f(ε)/4)τar/kλ

.

Proof. With the constant c5 appearing in Lemma 3, we set

c14(f, ε) := max0≤ε≤1

[(2/c5)(1 + f(ε)/3)τ ]1/r

and m0 := bc14(f, ε)c. Then

(c5/2)mr ≥ (1 + f(ε)/3)τ for m > m0. (35)

By (28) and Lemma 8, we have

P(Σ(Z0) ∈ a(1, 1 + h), ϑ(Z0) ≥ ε) =∑m∈N

qa,ε,h(m)

≤ c13(σ0)har/kλ

(m0∑

m=1

mrqa,ε,1(m) +∑

m>m0

mrqa,ε,1(m)

).

For the estimation of qa,ε,1(m) we use Lemma 5 for m ≤ m0 and Lemma 3 for m > m0,observing that qa,ε,1(m) ≤ qa,0,1(m). Then we can continue in the same way as in the proofof Proposition 7.1 in [4], where [4, (24)] is replaced by (35).

The proof of Theorem 1 can now be completed in exactly the same way as the proof ofTheorem 1 in [4]. The latter used only Lemma 3.2 and Proposition 7.1 of [4], and our presentLemmas 1 and 9 have the same structure as those results; they differ only in some parameters.In Lemma 9, we may assume σ0 = 1, then we obtain Theorem 1 under the assumption thatar/kλ ≥ 1. If we choose the constant c in (8) so that c > expc0f(ε), then (8) holds generally.

7 A Case of Large Cells with Indeterminate Shape

In this section, we want to study the situation of Theorem 1 in the special case where the sizefunctional Σ is equal to the parameter functional Φ; we will then call Φ(K) the Φ-content ofK. For a stationary and isotropic Poisson hyperplane process, this corresponds to the casewhere the size functional is essentially the mean width, the case excluded in [5, Theorem2]. The inequality (4) is now a tautological equality. Hence, every convex body K ∈ Kd

o

is an extremal body, and thus every deviation functional ϑ for Φ and Σ is identically zero.Therefore, P(ϑ(Z0) ≥ ε | Φ(Z0) ∈ [a, b)) = 0 for every ε > 0, so that inequality (8) is satisfiedtrivially, and provides no information.

26

That, in fact, the condition of large Φ-content need not influence the shape of the zerocell, can at least be seen in a very special case. Møller and Zuyev [11] have shown (ex-tending a result of Miles), that under the condition fd−1(Z0) = N , the Φ-content and theshape of the zero cell are stochastically independent. Now let X be the stationary Poissonhyperplane process where the directional distribution is concentrated, with equal masses, at±e1, . . . ,±ed, for an orthonormal basis (e1, . . . , ed) of Rd (a ‘cuboid process’, as studied in[2]). Then fd−1(Z0) = 2d with probability 1. It follows that, in this case, the Φ-content ofthe zero cell and its shape are stochastically independent.

The latter independence of shape and size does not hold for more general Poisson hy-perplane processes, but the following result on asymptotic independence can be proved. Itshows that no limit shape of Z0 with respect to Φ exists if the directional distribution hasfinite support.

Theorem 4. Suppose that the directional distribution ϕ of the hyperplane process X hasfinite support consisting of N points. Then, for every Borel set A ⊂ sD(Kd

o),

lima→∞

P(sD(Z0) ∈ A | Φ(Z0) ≥ a) = P(sD(Z0) ∈ A | fd−1(Z0) = N).

Proof. Let A be a Borel set in sD(Kdo). In the following, we define sD(A) := αA : α > 0

also for unbounded convex sets A ⊂ Rd, so that sD(P (H(n))) ∈ A implies that P (H(n)) isbounded. For n ∈ N and a > 0, we define

R(A, n) :=

(H1, . . . ,Hn) ∈ (Hd)n : sD(P (H(n))) ∈ A, fd−1(P (H(n))) = n

,

R(A, n, a) := (H1, . . . ,Hn) ∈ R(A, n) : Φ(P (H(n))) ≥ a,

S(A, n) :=

(H1, . . . ,Hn−1,u) ∈ (Hd)n−1 × Sd−1 : (H1, . . . ,Hn−1,H(u, 1)) ∈ R(A, n)

.

From Lemma 7, with B := K ∈ Kϕ : sD(K) ∈ A, Φ(K) ≥ a, we get, similarly as in theproof of Lemma 8,

P(sD(Z0) ∈ A, Φ(Z0) ≥ a, fd−1(Z0) = n)

=λn

n!

∫R(A,n,a)

exp−Φ(P (H(n)))λµn(d(H1, . . . ,Hn))

=λn

n!

∫Sd−1

. . .

∫Sd−1

∫ ∞

0. . .

∫ ∞

01(H(u1, t1), . . . ,H(un, tn)) ∈ R(A, n, a)

× exp−Φ(H−(u1, t1) ∩ · · · ∩H−(un, tn))λ

× (t1 · · · tn)r−1 dt1 · · ·dtn ϕ(du1) · · ·ϕ(dun)

=λn

n!

∫(Hd)n−1

∫Sd−1

1(H1, . . . ,Hn−1,H(u, 1)) ∈ R(A, n)

×∫ ∞

01zrΦ(H−

1 ∩ · · · ∩H−n−1 ∩H−(u, 1)) ≥ a

× exp

−zrΦ(H−

1 ∩ · · · ∩H−n−1 ∩H−(u, 1))λ

zrn−1 dz

×ϕ(du) µn−1(d(H1, . . . ,Hn−1)).

27

To compute the inner integral∫∞0 (· · · )dz, we fix (H1, . . . ,Hn−1,u) ∈ S(A, n), write

H−1 ∩ · · · ∩H−

n−1 ∩H−(u, 1) =: P

and define za = za(H1, . . . ,Hn−1,u) by Φ(zaP ) = a. Substituting z = zas, we get∫ ∞

0(· · · ) dz =

∫ ∞

01Φ(zP ) ≥ a exp−Φ(zP )λzrn−1 dz

=∫ ∞

za

exp−Φ(zP )λzrn−1 dz

=∫ ∞

1exp−sraλzrn

a srn−1 ds.

This yields

P(sD(Z0) ∈ A, Φ(Z0) ≥ a, fd−1(Z0) = n)

=λn

n!

∫(Hd)n−1

∫Sd−1

1(H1, . . . ,Hn−1,u) ∈ S(A, n)∫ ∞

1exp−sraλzrn

a srn−1 ds

×ϕ(du) µn−1(d(H1, . . . ,Hn−1))

=λn

n!

∫ ∞

1exp−sraλsrn−1 ds

×∫

S(A,n)za(H1, . . . ,Hn−1,u)rn ϕ(du) µn−1(d(H1, . . . ,Hn−1)).

Here we substitute sraλ = x, to obtain∫ ∞

1exp−sraλsrn−1 ds =

1r(aλ)−n

∫ ∞

aλe−xxn−1 dx.

Observing that zrna /an = zrn

1 , we get

P(sD(Z0) ∈ A, Φ(Z0) ≥ a, fd−1(Z0) = n) =1

rn!

∫ ∞

aλe−xxn−1 dx · I(A, n) (36)

withI(A, n) :=

∫S(A,n)

z1(H1, . . . ,Hn−1,u)rn ϕ(du) µn−1(d(H1, . . . ,Hn−1)).

Letting a tend to zero, we see that (36) holds also for a = 0.We remark that a consequence of (36), namely

P(Φ(Z0) ≥ a | fd−1(Z0) = n, sD(Z0) ∈ A) =1n!

∫ ∞

aλe−xxn−1 dx

for the sets A with P(fd−1(Z0) = n, sD(Z0) ∈ A) > 0, could also be deduced from the workof Møller and Zuyev [11]. We wanted to give here the explicit form of I(A, n), in view of theremark at the end of this section.

By assumption, the support of ϕ has N points. Therefore,

P(sD(Z0) ∈ A | Φ(Z0) ≥ a) =

N∑n=d+1

1n!

∫ ∞

aλe−xxn−1 dx · I(A, n)

N∑n=d+1

1n!

∫ ∞

aλe−xxn−1 dx · I(sD(Kd

o), n)

.

28

Since ∫ ∞

aλe−xxn−1 dx = e−aλ(aλ)n−1(1 + o(1)) as a →∞,

we get

lima→∞

P(sD(Z0) ∈ A | Φ(Z0) ≥ a) =I(A, N)

I(sD(Kdo), N)

= P(sD(Z0) ∈ A | fd−1(Z0) = N).

Theorem 4 shows clearly that asymptotically the condition of large Φ-content has little in-fluence on the shape of the zero cell. In fact, if N > d + 1, we can construct quite differentlylooking Borel sets A of D-shapes in Kϕ with N facets for which I(A, N) > 0. Hence,

lima→∞

P(sD(Z0) ∈ A | Φ(Z0) ≥ a) > 0

can be satisfied for essentially different sets A, which is in stark contrast to the existence oflimit shapes.

8 Proof of Theorem 2

Let κ ∈ (0, 1) and m ∈ N, and suppose that ar/kλ ≥ σ0 with some constant σ0 > 0. Let Ba

and C be chosen as in the proof of Lemma 5. Suppose that H1, . . . ,HN ∈ HC are such thatP (H(N)) ∈ Ka,0,1(m); then

Φ(P (H(N))) ≥ Φ(Ba) (37)

by (31) (which holds also for ε = 0). As in the proof of Lemma 5, there are ν = ν(d, ϕ, r, κ)vertices of P (H(N)) such that the convex hull Q = Q(P (H(N))) of these vertices satisfies

Φ(Q) ≥ (1− κ/8)Φ(Ba), (38)

by (32) and (37). The proof of Lemma 5 (using (38) instead of (32)) shows how this leads to

qa,0,1(m) ≤ c16(κ)[(ar/kλ)dν + 1

]mrdν exp

−(1− κ/8)τar/kλ

≤ c17(κ)mrdν exp

−(1− κ/4)τar/kλ

. (39)

Now we can argue as in the proof of Lemma 9. Let h ∈ (0, 1/2). We set

c18(κ) := [(2/c5)(1− κ/4)τ ]1/r

and m0 := bc18(κ)c. Then

(c5/2)mr ≥ (1− κ/4)τ for m > m0. (40)

By (28) and Lemma 8, both for ε = 0, we have

P(Σ(Z0) ∈ a(1, 1 + h)) =∑m∈N

qa,0,h(m)

≤ c13(σ0)har/kλ

(m0∑

m=1

mrqa,0,1(m) +∑

m>m0

mrqa,0,1(m)

).

29

For the estimation of qa,0,1(m) we use (39) for m ≤ m0 and Lemma 3 for m > m0. Wecontinue as in the proof of Proposition 7.1 in [4], where [4, (24)] is now replaced by (40). Weconclude that

P(Σ(Z0) ∈ a(1, 1 + h)) ≤ c19(κ, σ0)h exp−(1− κ/2)τar/kλ

. (41)

Here we replace a by sia with 1 < s < 1 + h and i ∈ N0 and estimate the exponent by

(1− κ/2)τsir/kar/kλ = (1− κ)τsir/kar/kλ + (κ/2)τsir/kar/kλ

≥ (1− κ)τar/kλ + (κ/2)τsir/kσ0.

Since (a,∞) =⋃∞

i=0 sia(1, 1 + h) (and P(Σ(Z0) = a) = 0), we get (using h < 1/2)

P(Σ(Z0) ≥ a) ≤∞∑i=0

P(Σ(Z0) ∈ sia(1, 1 + h))

≤∞∑i=0

c19(κ, σ0) exp−(1− κ/2)τsir/kar/kλ

≤ c19(κ, σ0) exp−(1− κ)τar/kλ

∞∑i=0

exp−(κ/2)τsir/kσ0

≤ c20(κ, σ0, s) exp

−(1− κ)τar/kλ

.

Thus, together with (19), we have

exp−τar/kλ

≤ P(Σ(Z0) ≥ a) ≤ c20(κ, σ0, s) exp

−(1− κ)τar/kλ

.

This yieldslim infa→∞

a−r/k ln P(Σ(Z0) ≥ a) ≥ −τλ

andlim sup

a→∞a−r/k ln P(Σ(Z0) ≥ a) ≤ −(1− κ)τλ.

Here the left-hand side is independent of κ, hence we conclude that

lima→∞

a−r/k ln P(Σ(Z0) ≥ a) = −τλ.

This completes the proof of Theorem 2.

9 Proof of Inequality (9)

Now we prove the inequality (9). From (41) we can deduce that the distribution of Σ(Z0)is absolutely continuous with respect to the Lebesgue measure λ1 on R. In fact, we haveP(Σ(Z0) = 0) = 0. If a set M ⊂ [a,∞) with ar/kλ ≥ σ0 for some σ0 > 0 is covered bycountably many intervals of total length ε, then it follows from (41) that the sum of thePΣ(Z0)-measures of these intervals is at most c21ε, with a constant c21 not depending on ε.From this, the absolute continuity of PΣ(Z0) with respect to λ1 follows. Moreover, Lemma

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1 with β = 1, say, shows that the Radon–Nikodym derivative of PΣ(Z0) with respect to λ1

satisfies

dPΣ(Z0)

dλ1(a) = lim

h↓0

P(Σ(Z0) ∈ a(1, 1 + h))ah

≥ c1a(ar/kλ)N exp−2τar/kλ > 0,

for λ1-almost all a > 0. Hence PΣ(Z0) and λ1 are equivalent measures. Since

P(ϑ(Z0) ≥ ε, Σ(Z0) ∈ B) =∫

BP(ϑ(Z0) ≥ ε | Σ(Z0) = a) PΣ(Z0)(da)

for any Borel set B ⊂ R, we obtain from Lebesgue’s differentiation theorem that

P(ϑ(Z0) ≥ ε | Σ(Z0) = a) = limh↓0

P(ϑ(Z0) ≥ ε, Σ(Z0) ∈ a(1, 1 + h))P(Σ(Z0) ∈ a(1, 1 + h))

for λ1-almost all a > 0. Here we use Lemma 9, with σ0 = 1, for the upper estimation ofthe numerator and Lemma 1 for the lower estimation of the denominator, and thus deduceinequality (9), for ar/kλ ≥ 1. Then we adapt the constant c as in the proof of Theorem 1 toobtain (9) for all a > 0.

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[3] Goldman, A., Sur une conjecture de D.G. Kendall concernant la cellule de Crofton duplan et sur sa contrepartie brownienne. Ann. Probab. 26 (1998), 1727–1750.

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[7] Kovalenko, I.N., An extension of a conjecture of D.G. Kendall concerning shapes ofrandom polygons to Poisson Voronoı cells. In: Engel, P. et al. (eds.), Voronoı’s impacton modern science. Book I. Transl. from the Ukrainian. Kyiv: Institute of Mathematics.Proc. Inst. Math. Natl. Acad. Sci. Ukr., Math. Appl. 212 (1998), 266–274.

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[10] Miles, R.E., A heuristic proof of a long-standing conjecture of D.G. Kendall concerningthe shapes of certain large random polygons. Adv. Appl. Prob. 27 (1995), 397–417.

[11] Møller, J. and Zuyev, S., Gamma-type results and other related properties of Poissonprocesses. Adv. Appl. Prob. 28 (1996), 662–673.

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Daniel Hug, Rolf SchneiderMathematisches InstitutAlbert-Ludwigs-UniversitatD-79104 Freiburg i.Br.Germany

daniel.hug@math.uni-freiburg.derolf.schneider@math.uni-freiburg.de

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